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Talk:Hyperfactorial
Okay, some thoughts concerning this function. # Are there any numbers equal to the sum of the hyperfactorials of their digits? A brief brute-force search convinces me that there are no such numbers; if there are any, they're very large. # Can we generalize H''(''n) for the real numbers or the complex plane, the way the gamma function generalizes the factorial? # Is there an equivalent of Stirling's approximation? I figured out \ln H(n) - (n \ln n) / 2 \approx \frac{1}{4}n^2 (2 \ln n - 1) , but I'm a beginner at integral calculus and this could be totally wrong. FB100Z • talk • 19:33, September 6, 2012 (UTC) Okay, I've got the approximation H(n) \approx n^{n^2/2 + n/2}e^{-n^2/4} . It seems pretty accurate, but it's hard to tell with numbers this size. It looks vaguely like the integrand of \Gamma(n) — could this lead to our continuous hyperfactorial? FB100Z • talk • 21:21, September 6, 2012 (UTC) :This one? Hyperfactorials and the K-function --Cloudy176 (talk) 13:08, September 10, 2012 (UTC) ::D'oh, didn't see that. I'm smart. FB100Z • talk • 21:53, September 10, 2012 (UTC) :About the first question: If such numbers exist then it should be less than 10^{36} , because if n is a d-digit number satisfying the conditions, then 10^{d-1} < n < d \cdot H(9) , and 37 H(9) \approx 7.9838 \times 10^{35} < 10^{36} . --Cloudy176 (talk) 13:28, September 10, 2012 (UTC) :: is obviously too big for brute force, so I'll have to use a sneakier method (maybe building from the digits up). I'll look into this. FB100Z • talk • 21:53, September 10, 2012 (UTC) :::Rather than search through all numbers between 0 and 10^{36} , you can instead search through all combinations of digits of 36 or less, and then check if the sum of the hyperfactorials of those digits has the right combination of digits. The total number of such combinations is {36 + 10 - 1 \choose 10 - 1} = 886163135 , so it's possible to search through.Deedlit11 (talk) 13:22, September 26, 2012 (UTC) 2 digits Runtime: 0s 3 digits Runtime: 0s 4 digits Runtime: 0s 5 digits Runtime: 0s 6 digits Runtime: 0s 7 digits Runtime: 0s 8 digits Runtime: 1s 9 digits Runtime: 1s 10 digits Runtime: 3s 11 digits Runtime: 5s 12 digits Runtime: 9s 13 digits Runtime: 15s 14 digits Runtime: 26s 15 digits Runtime: 41s This might take a while, folks. FB100Z • talk • 20:09, September 27, 2012 (UTC) :Got a few optimizations that shaved 21 digits down to a minute, and working on more. Much to my chagrin, my estimate for total computation time came up to 8-12 hours. :Question: Given a digit sequence , , , , ..., with each digit greater than or equal to the last, is there a quick way to ensure that s'' = H( ) + H( ) + H( ) + H( ) + ... + H( ) (the sum of their hyperfactorials) has n digits without actually computing the sum? FB100Z • talk • 21:12, September 27, 2012 (UTC) Hyperderivative The hyperfactorial opens up to some useless analogs of common analytical functions. Define the h-exponential: \= 1 + x + \frac{x^2}{H(2)} + \frac{x^3}{H(3)} + \ldots\ and the h-trig functions \begin{eqnarray} \cos_H(x) &=& \frac{e_H^{ix} + e_H^{-ix}}{2} \\ \sin_H(x) &=& \frac{e_H^{ix} - e_H^{-ix}}{2i} \\ \tan_H(x) &=& \frac{\sin_H(x)}{\cos_H(x)} \\ \end{eqnarray} FB100Z • talk • 20:00, June 6, 2013 (UTC)